electrical loads the first step in the electrical designing for any construction is to estimate the...

electrical Loads

• the first step in the electrical designing for any construction is to estimate the electrical loads

• This enables the electrical engineering (designer) to know exactly who should the electrical power feed this construction

An-Najah National UniversityFaculty of Engineering

Electrical Engineering Department

Electrical Loads

• Electrical loads in the buildings can be divided into three types:- Electrical loads used in lighting ( illumination system) , different types of lamps) Electrical loads feed the normal sockets and outlets ( used for small electrical devices , TV, radio , washing machine, etc….) Electrical loads that feed the mechanical devices used in the buildings ( elevators , heating system ,cooling system , pumps ,etc)



estimation of the electrical loads used in lighting

• the estimation of the electrical loads used in lighting is achieved by calculating the sum of the lamps used in the illumination system • there are different types of lamps :- Incandescent Lamp

Gas-discharge lamps



Old method for Electrical Load estimation in lightings

• this methods assumes that all the lighting system consists of either Incandescent Lamp or florescent lamps or a mixed of them . • in the case of using incandescent lamps :- Power(watt/m2) = illuminances (Lux)/2.78 • in the case of using incandescent lamps :- Power(watt/m2) = illuminances (Lux)/11.148 form the above equations it is quite clear that the using of florescent lamps can save power so it is quite important to use them as much as possible.This method is no longer used as it depends on approximations in addition to the fact that there a lot of different types of lamps used these days



example

• An office with an area of 60 m2, and the illuminance equals 500 lux, calculate the power needed in the cases of using the incandescent lamp and florescent lamps.

Answer :- • in the case of using candescent lamps :- P= 60* 500 /2.78709 = 10763 watt• in the case of using florescent lamps : P= 60* 500 /11.148 = 2691 watt power saving using the florescent lamps= 10763-2691= 8072 watt



Unit or Specific Load per Square meter•To over come the disadvantage of the previous method the “Unit or specific load per square meter” which depends on the natural using of the space in the building.•This method give more accurate results than the previous one.



Room Type Unit or specific load(watt/m2)

Living room (not dark colours for walls) 20-30

Living room (dark colours for walls) 40-50

children room (not dark colours for walls) 30-40

children room (dark colours for walls) 60-70

Sleeping room 20-30

kitchen 50-60

Working areas 40-50

Stairs , corridors, 20-30



Room Type Unit or specific load(watt/m2)

theatres 10

banks 21

Hair dressing shops 32

Worship areas 10

clubs 21

hospitals 21

hotels 21

offices 53

schools 32

Restaurants 21

ways for energy consumption saving in lights systems

• using lamps unites with high efficiency which is called “energy saving lamps” which have a high efficiency per watt in addition to the fact that they lasting for long times compared to the normal lamps• using lamps unites with low electrical losses on the form of heat which reduce the need of using the cooling systems for the construction.• reducing the light losses by using lamp units with low light losses coefficient• using lamp units with higher coefficient utilization



Unit Power density procedure

• this method differs than the previous methods in:- 1- it takes into account the shape of the rooms or the space(square,rectangle, etc) 2- it assumes that the lighting system in that room or space is based on the maximum using of the electrical power.

This method depends on dividing the building into rooms depending on the nature of using. And for every room or space there is a tables shows base power unit density(UPD) for that space in watt/m2




• the power needed for lighting a room or a space according to this method is:-

Power= Area* Power unit density(UPD)*room factor* space utilization factor(SUF). Power= A* a1* a2 Power: wattArea (A):m2

Power Unit density(UPD): (F):watt/ m2

Room factor : (a1)Space utilization factor :(a2)




Power Unit density(UPD): obtained from tables(1.2)



Room Type Power unity density

Bank 50.59

hospitals 15.07

offices 34.44

buildings 23.68

hotels 15.07

Libraries 9.69

Schools 23.68

Shops 40.9



Unit Power density procedureRoom factor : (a1)

Room factor (a1) is a number between 1.0 and 2.0 and its important comes from that it shows the effect of the room or space shape on the lighting system . From table (2.2)

Eg 1 :- length of the room is 2.4 . Width is 3.7 and the length is 2.7. the room factor from the table will be 1.8

Eg 2 :- length of the room is 7.3 . Width is 2.4 and the length is 3.4. the room factor from the table will be 1.9

If the shape of the room is not regular we choose a nearest regular shape such that the area of the regular shape is equal to the irregular shape of the room or space



Unit Power density procedureSpace utilization factor :(a2)

The value of the space utilization factor (S.U.F) is between 0.4 and 1.0 .

it is depends on the ration between the area of the room or space that is used and the total area of the space or the room.

Ratio between the used area and the total area space utilization factor (S.U.F)

0.5-1.0 1

0.4-0.49 0.85

0.3-0.39 0.77

0.2-0.29 0.55

Up to 0.19 0.4




• After calculating the electrical power needed for lighting the rooms it means that we calculate the electrical power needed for lighting all the listed areas.

•For the unlisted spaces . First : the area of these spaces are calculated and it equals the area of each flat in the building –the area of the listed spaces.

•The electrical power for the unlisted spaces is then calculated by multiplying its area by 2.15 watt/m2

•the total electrical power for lighting the inner area will be sum of the powers needed for listed and unlisted area.

Unlisted areas




• it consists of 1- the electrical power calculated for the inner space of the project(listed and un listed) 2- the electrical power needed for lighting the outside of the project

a- the electrical power needed for lighting the outside walls of the project (5% of the electrical power needed for the inner lighting ) b- the entrance of the building and the exit (table 1.2)

Unit power density for the inner lighting of the project= power for inner light/ total area of the flats Unit power density for the lighting of the project= power for inner light/ total area of the project

The maximum electrical power for the whole building




• after calculating the total power for the whole building we multiply this power by factor called demand factor which equals 0.85 in the case of lighting.

Unit power density for the inner lighting of the project= power for inner light/ total area of the flats Unit power density for the lighting of the project= power for inner light/ total area of the project

The maximum electrical power for the whole building



Electrical loads needed for feeding the sockets and outlets

• used to feed electrical devices with small power

• if the devices that connected with socket is already known, then the rated power for that device is considered.• if not , then for each socket , we assume that the power for the devices that connected to that socket is between 200 to 250 watt if it is single•If it is double socket then we assume the power for that socket is about 350 watt.



Electrical loads needed for feeding the sockets and outlets

•Table 5.1 shows the number of sockets needed in each room in the building

•Form the table the number of sockets needed in kitchen is 4 in addition to a special socket for the cooker unit.

•The total power needed for feeding the sockets will be then multiply by 0.7 as a demand factor.



Electrical loads needed for feeding the mechanical devices

•Conditiong unit :- each ton refrigeration needs 1500 watt electrical power.•Table 6.1 shows the number of ton refrigerating needed in different types of buildings•For the pumps :- the electrical engineering obtained the mechanical Horsepower (HP) needed for each pumps

•The electrical power =for that pump will be: mechanical Horsepower *746 watt

•For the Escalator and Electrical Stair the power will be given from the catalogue



Electrical lamps

• Electrical lamps that used in lighting is divided into two main type:- 1- Incandescent lamps 2- Discharge lamps

For the incandescent lamps :- the light is emitted as a result of electrical current passing through a filament , which makes it heated to a high temperature then glowing and luminating .

For the discharge lamps:- the light is emitted due to the glowing of the gas atoms between the poles of the lamp.



Electrical lamps

• Specification of electrical lamps:-

1- Luminous flux (lumen) 2- Luminous efficiency ( Lumen/watt) 3- Life time 4- Illumination (Lux) or (Lumen/m2) 5- power , voltage (watt ,V) 6- size of the lamp

The luminous efficacy of a lamp



Electrical lamps

• Incandescent lamps

1 – Filament (Tungstun) 2- Bulb 3- Base 4- inert gas or empty volume




Electrical lamps

Incandescent lamp


• the current is fed from the electrical source to the filament using the base• when the current pass through a filament, its temperature is increased and after a short time it is becomes glowing and emitting the light.

• carbon used to be used in the filament , but now a days , Tungstun is used in filament and sometimes some elements such as AL, K, SI are added to Tunstun to improve the hardness of the filament

• luminous efficiency of the incandescent lamps increases as power of that lamps increase . For example, a lamp with power 150 watt give more illumination in a percentage of 34% than using three 50 watt lamps.



Electrical lamps

Incandescent lamp


• the tungstun incandescent lamps are used a lot because of:-

1- it is found in a lot of shapes (table 2.3)

2- the quality of light (suitable for the eye)

3- is very cheap

•The lifetime for the incandescent is about 1000 hour, •The luminous flux of the incandescent lamps is directly related to the applied voltage on the lamp. •Table 3.3 shows the relation between the luminous flux , voltage, and the lifetime .



Electrical lamps

Incandescent lamp


•The end life of the Tungstun lamps

•The melting of the filament is the indication of the end life of the Tungstun incandescent lamps•Due to a defect in a position on the filament , a hot-spot is formed on it, and the temperature of the on that position at beginning will be higher than other position on the filament.• the high temperature makes the Tungstun to evaporate.

•All these process happened at the beginning of the lighting .•The evaporated Tungstun is stuck on the inner surface of the bulb making a black colour to appears on the bulb surface.(blackening).



Electrical lamps

Incandescent lamp


•The end life of the Tungstun lamps•The blackening phenomena appears only in the Tungstun lamps where the bulb is evacuated from the air

• To overcome the blackening phenomena , a Halogen Lamps are used where one of the following halogen is used (Cl, Br, I, F) in addition to the inert gas. This halogen helps to return back the evaporated Tungstun. •they are more efficient than incandescent bulbs using only half the energy to produce the same light output and last twice as long•The halogen lamps are used for specific applications where a low voltage and clear light is needed like , cinema , projectors, theaters, cars lamps , in TVs



Electrical lamps

Discharge lamps


• the principal operation of the discharge lamps is based on an electrical discharge in the atoms of the inert gas or the vapour of metals or a mixed of them.• this results on some visible lights •It is divided into two types:- 1- low pressure discharge lamps (Neon , Florescent) 2- high pressure discharge lamps (Soduim and mercury )



Electrical lamps Florescent lamps

• The current passes through the circuit heating up the filament in each electrode, which are located at both ends of the tube. •The heated electrodes in addition to the high voltage applied to two ends of the lamps cauasing electrical discharge (ionizing the gas(argon)). •The mercury vapor becomes "excited" and it generates radiant energy, mainly in the ultraviolet range. •This energy causes the phosphor coating on the inside of the tube to fluorescent, converting the ultraviolet into visible light.



Electrical lamps Florescent lamps

• in order to obtain a light from florescent lamps, the following things should be happened 1- electrical discharge 2- tranforming the ultraviolet radiated energy to a visible light



Electrical lamps

Power of the lamps(watt)

Incandescent lamps 1- general service lamps (15 to 250 watt) 2- projectors (40-1000) watt 3- halogen Lamps (1000-2000) watt 4- table lamps for specific use (15- 50)watt (positional use)Florescent lamps 1- general (15-80) watt 2- special types (125-200) watt 3- table lamps (4-31) watt (positional use)



Electrical lamps

Mercury Lamps 1- general (80-1000) watt

Soduim Lamps 1- low pressure (35-180) watt

2- high pressure (70-1000) watt



Electrical lamps

Lifetime

Incandescent Lamps :-

1- Genral (1000 ) hour 2- projector (750-1500) hour 3- Halogen (2000) hourFlorescent :- 1- (15-80 ) watt :- 1000 hour 2- (125-200) watt:- 3000 hour 3- (4-13) watt:- 2000 hourMercurry :-7500 hour Soduim :- 2000 hour

CFL :- (5000 -15000)



Electrical lamps

Luminous efficiency ( Lumen/watt)

Lamp type Luminous efficiency

Incandescent 25

Florescent 80

Mercury 60

Soduim (Low pressure) 185

Soduim (high pressure) 140

CFL (sl, pl) (compact florescent lamps) (energy saving lamps)

75-95



Electrical lamps

Energy saving Lamps (CFL)

It is new lamps and comes to replace the traditional lamps (florescent and incandescent)

It has the following characteristics:-

1- Low consumption of energy2- long life time (5000 hour)3- high luminous efficiency (50 lumen/watt)4- have a small size.



Electrical lamps

Comparison between incandescent lamps and CFL lamps

1- Life time 5- Size * incandescent 1000 hour * bigger * CFL 5000 hour * smaller2- power 6- weight * incandescent :75 watt * bigger * CFL : 18 watt * smaller3- luminous flux :- 7- cost * incandescent : 900 lumen * less * CFL : 90 lumen * more4- luminous efficiency * incandescent : 12 lumen/watt * CFL 50 lumen/watt



Electrical lamps

Comparison between incandescent lamps and CFL lamps

Eg:- compare between incandescent lamps (75 watt) and CFL lamp with power (18 ) watt . If The price of incandescent is 0.200 JD and the price of the florescent is 4.5 JD and the price of the kwh is 0.04 JD assuming that both kamps will work 5000 hourNo of incandescent lamps =5000/1000=5No if CFL lamps =1The price of the incandescent lamps =5*0.2 = 1 JDThe price of the CFL lamps = 4.5 JDThe power consumed by incandescent in 5000 hour in Kwh= 5000 *75/1000 = 375 kwhThe price of the kwh = 375 * 0.04 = 15 JD Total for incandescent = 15+1= 16 JDKwh for CFL = 5000*18/1000= 90Price of kwh = 90 *0.04= 3.6 JD

Total price for CFL = 3.6+4.5 = 8.1 JDSaving = 16- 8.1 = 7.9 JD



Lighting calculations

The objective of performing the lighting calculations is to determine the type, power and Distribution the lamps unites

This needs the to know the illumination needed in the sapce depinding on the natural use of the space

the method that we want to use here is called Lumen method




Lumen Method

This method depends on using the utilization factor assuming that the illumination is distributedRegularly in all directions in the space.

N= Em.A/n*Fl*ku*kn

N= no of lamp unites needed to obtain the desired illuminationEm= illumination (obtained from tablesA = arean= no of lamps in the uniteFl= luminous flux for the lamp (lumen)Ku= utilization factorKn= maintenance factor




Lumen Method

Kn= maintenance factor

The maintenance factor depends on the situation of space :-

It is value obtained from the table below:-

Room situation Maintenance factor

Normal 0.8

Dirty 0.7

Very dirty 0.6




Lumen Method

Ku= utilization factor The utilization factor depends on the dimensions of the room .

Kr Utilization factor %

0.6 35

0.8 44

1 51

1.25 58

1.5 64

2 72

2.5 77

3 81

4 85

5 89

>5 90




Lumen Method

Ku= utilization factor Where kr is the room index and it is calculated as follows:-

Kr= L*w/Hm(L+w)Where

Kr:- room indexL :- length of the roomw:- width of the roomHm:- the high of the lamp unit from the surface of works in m




Lumen Methodkr

If the height of the room is H , and the height of the woorking surface from the ground of the room is Hp, and the lamp unit is hanged a distance HL from the ceiling then Hm=H-HL-Hp

Usually Hp = 75 cm except it is given a value different than that.




Lumen Methodeg

an office wit dimensions = 8m length , 5 m width and 3 m height . Calculate the number of unit lamps needed for illumination.

Solution:- we use lumilux flourescent lamps from table 16.3 page 111 where luminous flux = Fl=L=3450 lumen .Ilumination =500 lux. Page 275.Maintenance factor=kn= 0.8Utilization factpr kr= L*w/Hm(L+w).HL=0Hp=0.75 mHm=3-0-0.75=2.25 mKr=8*5/(8+5)*2.25=1.37We choose kr=0.58+0.64/2=0.61N=500*8*5/2*3450*0.61*0.8=5.59 = 6 units



Electrical Conductors

• the electrical conductors that are used in electrical installation are usually made from AL or CU or a mixed of them.

•The difference between the AL and CU is that the specific resistance for the Cu is less than that in AL

•This means that a cable with bigger radius is needed in the case of AL compared to Cu to conduct the same current.




Conductor resistances and the factors that affect it

•Conductors have a high conductivity due to the existence of a high no of free electronsor in other meanings its electrical resistance is very low :- ρ=1/σWhere ρ is the resistivity and σ is the conductivity of the conductors.•The resisivity is an indication of the quality of the materials as the resistivity is wanted to be as small as possible .




The resistivity of the conductors is affected by the temperature according to the following formulas:-

ρt=ρ0(1+ℓ (T-T0)

Whereρt is the resistivity at temperature Tρ0 is the resistivity at 250

ℓ is the thermal expansion coefficient for the conductors at 250




thermal expansion coefficient for the conductors

Material Thermal expansion coefficient

silver 0.0038

Copper 0.00393

Aluminium 0.00377

Nickel 0.006

Iron 0.006

Platinum 0.0025

NI mixture 0.0001

http://www.4shared.com/file/_qrtIq3h/Pokemon_Platinum__U__English_V.html




The resistance of any cable is given as:-

R= ρL/AWhere R is the resistanceρ resistivityL length of the cableA area of the cableFrom the above equations the resistance depends on:-1-the conductor type (AL ,Cu)2- length of the cable3- area of the cable4- temperatureIt is meant by temperature :-1- the temperature of the space around the conductors2- the temperature of the cable itself which is produced by the current passing in the cable




Ampacity is defined as the maximum amount of electrical current which a cable can carry before sustaining immediate or progressive deterioration

1- its insulation temperature rating;

2- the electrical resistance of the cable material;

3-frequency of the current, in the case of alternating current;

4 - ability to dissipate heat, which depends on cable geometry and its surroundings;

5- ambient temperature.

Ampacity of cables

http://en.wikipedia.org/wiki/Thermal_insulation

http://en.wikipedia.org/wiki/Temperature




How to calculate the cross section area of the cable

1- calculate the designing current in the circuit ore part of the circuit2- according to that , the protection for the circuit is selected3- the choosing of the cross sectional area of the cable will be according to

I cable > I protection > I design




Coloures for electrical conductors

Current IEC

Protective earth (PE) Green/yellow bi-colour

Neutral (N) Blue

Single phase: Live (L)Three phase: L1 Brown

Three phase: L2 Black

Three phase: L3 Grey

http://en.wikipedia.org/wiki/Ground_and_neutral




Area of Conductor

Resistance at 20°C Max. ohm/km Current Rating (Amp.)

0.05 39.00 40.75 26.00 71.0 18.10 111.5 12.10 14

2.5 7.41 19

4.0 4.95 26

6.0 3.30 33

10.0 1.910 45

16.0 1.210 60

25.0 0.780 75

35.0 0.554 95

50.0 0.386 125




Drop voltage calculations

Voltage drop = V1 – V2 =2(I1*r1 +I2*r2+I3*r3+I4*r4)I1,2,3,4 are the currentsr1,2,3,4 are the branches resistances

Let us assume I1=i1+i2+i3+i4 I2=i2+i3+i4 I3=i3+i4 I4= i4




Drop voltage calculations

Let us assume R1=r1+r2+r3+r4 R2=r2+r3+r4 R3=r3+r4 R4= r4

Voltage drop = 2(i1*R1+i2*R2+i3*R3+i4*R4)=2∑iαR α

As R=ρL/AVoltage drop = 2 ρ/A ∑ILA= (2 ρ/voltage drop) ∑IL



Sockets connections

220 V -50 Hz

N P



Switch connections (2 lamps in series with single switch

220 V -50 Hz

N P



Switch connections (2 lamps in parallel with single switch 220 V -50 Hz

N P



Switch connections (2 lamps with double switch

N P

220 V -50 Hz



Switch connections (2 lamps with double switch

N E P

220 V -50 Hz

chandelier



Switch connections (2 lamps with double switch 220 V -50 Hz

E N L



Sockets connections

220 V -50 Hz

N E P



Sockets connections

N E P



lighting 2 lamps from 3 different places220 V -50 Hz

E N L



Bell with push button switch

N E P



Meausring devices

N E PV

A

Kwh

V A



Circuit breakersProtect from :-

1- short circuit 2- over load3- short circuit and over loadCurrent ratings of circuit breakers :

6 A, 10 A, 13 A, 16 A, 20 A, 25 A, 32 A, 40 A, 50 A, 63 A, 80 A and 100 Ainstantaneous tripping current, that is the minimum value of current that causes the circuit-breaker to trip without intentional time delay (i.e., in less than 100 ms), expressed in terms of In

Type Instantaneous tripping current B above 3 In up to and including 5 In

C above 5 In up to and including 10 In D above 10 In up to and including 20 In K above 8 In up to and including 12 In



Circuit breakersTypes of circuit breaker

1- thermal circuit breaker

2- magnetic circuit breaker

3- thermal and magnetic circuit breaker

In choosing circuit breakers we should consider the following

1- ICB>= maximum load current2- VCB> = V supply3- Ibreaking capacity > 1.2 IscMCB :- miniature circuit breaker MCCB :- moulded case circuit breaker (63AMP (min.))



Circuit breakers

No of units lamps in each feeder should be less than 9No of sockets in each feeder should be less than 5 if it is 2 A and less than 3 if it is 5 A socket

the circuit breaker for the lamp units feeder is 10 A and the cross section of wire is 1.5 mm2

The circuit breaker for the socket feeder is 16 A and the cross section of the wire is 2.5 mm2

There should be feeder for the kitchen and at least spare feeder



RCD single phase (residual-current device)

2* 40 A, 0.03 A

http://en.wikipedia.org/wiki/File:FISkizze.svg



RCD 3 phases (residual-current device)

4* 40 A, 0.03 A

http://en.wikipedia.org/wiki/File:Fi-rele2.gif



Many flats in building (main distribution board

KwH KwH KwH

3 phase supply




3 phase supply

KwHBasement 2

Basement 1

Ground floor

first floor

second floor

Third floor

fourth floor

DB-B2

DB-B1

DB-GF

DB-F1

DB-F2

DB-F3

DB-F4

5*16mm2 xlpe

5*16mm2 xlpe

5*16mm2 xlpe

5*16mm2 xlpe

5*16mm2 xlpe

5*16mm2 xlpe

5*16mm2 xlpe



Main DSB

3 * 160 A

KwH

DB-B2

5*16mm

2

3 * 63 A

DB-B15*16m

m2

3 * 63 A

DB-GF

5*16mm

2

3 * 63 A

DB-F1

5*16mm

2

3 * 63 A

DB-F2

5*16mm

2

3 * 63 A

DB-F3

5*16mm

2

3 * 63 A

DB-F4

5*16mm

2

3 * 63 A

From mains

RCD 160 A 30 mA



Basement 2

3 * 63 A

from DB-B2

light

3*1.5mm

2

10 A

light3*1.5m

m2

10 A

light

3*1.5mm

2

10 A

light

3*1.5mm

2

10 A

socket

3*2.5mm

2

16 A

socket

3*2.5mm

2

16 A

socket

3*2.5mm

2

16 A

spare

3*2.5mm

2

16 A

RCD 63 A 30 mA




3 phase supply

Basement 2

Basement 1

Ground floor

first floor

second floor

Third floor

fourth floor

DB-B2

DB-B1

DB-GF

DB-F1

DB-F2

DB-F3

DB-F4

3*16mm2 xlpe

3*16mm2 xlpe

3*16mm2 xlpe

3*16mm2 xlpe

3*16mm2 xlpe

3*16mm2 xlpe

3*16mm2 xlpe

KwH

KwH

KwH

KwH

KwH

KwH

KwH




3 phase supply

KwH

Basement 2

Basement 1

Ground floor

first floor

second floor

Third floor

fourth floor

DB-B2

DB-B1

DB-GF

DB-F1

DB-F2

DB-F3

DB-F4

3*16mm2 xlpe

3*16mm2 xlpe

3*16mm2 xlpe

3*16mm2 xlpe

3*16mm2 xlpe

3*16mm2 xlpe

KwH KwH KwH KwH KwHKwH



Main DSB

3 * 160 A

KwH

DB-B2

3*16mm

2

1 * 63 A

DB-B13*16m

m2

1 * 63 A

DB-GF

3*16mm

2

1 * 63 A

DB-F1

3*16mm

2

1 * 63 A

DB-F2

3*16mm

2

1 * 63 A

DB-F3

3*16mm

2

1 * 63 A

DB-F4

3*16mm

2

1 * 63 A

From mains

RCD 160 A 30 mA



Basement 2

1 * 63 A

from DB-B2

light

3*1.5mm

2

10 A

light3*1.5m

m2

10 A

light

3*1.5mm

2

10 A

light

3*1.5mm

2

10 A

socket

3*2.5mm

2

16 A

socket

3*2.5mm

2

16 A

socket

3*2.5mm

2

16 A

spare

3*2.5mm

2

16 A

RCD 63 A 30 mA



Electrical installation In factories

the electrical load in the factories depends mainly on the technological process used in the factoryAs the industrial process is determined in the planning design for the factory , it is possible to use the electrical information about the loads directly from the company that designs the machinesThe electrical installation in the factories consists of the following elements:-

1- main distribution board2- feeders3- auxiliary distribution boards and circuit breakers




connected with the main supply coming from the electrical distribution company and distributed from it the feeders to auxiliary distribution boards it consists of :-1- mean switch (circuit breaker) :- disconnect the distribution board from the electrical power and for protection it can be either air switch or oil siwtch2- bus-bars :- one bus-bar for each phase ,N ,E3- auxiliary switches4- instrumentation devices (power , A , V, f)

Main distribution boards



Electrical installation In factories Auxiliary distribution boards




there are different types of distribution boards either single phase or 3 phase and it can be designed to be used outside the factory or inside it in chosing the distribution boards . It is better to chose a place easy to be reached and in the middle of the electrical loads as this will reduce the material used in the installation and reduce the voltage dropsIt better that the electrical installation in the factories to be clear to make the maintenance easyThe cables and the sockets should be designed to carry the short circuit current for short time complete information for safty procedures and the electricla loads on eachD

General Notes about electrical installation in factories



Electrical distribution in factories

radial distribution

ring system voltage drop is less , more reliable

Both radial and Ring system



Electrical distribution in factories

it is recommended to choose the substation to be near the centre of electrical loads in order to reduce the cost of the electrical netwrok and the no of circuit breaker and protection devices.

substations in factories

y(m)

x(m)

load 2 load 1

load 3

Xo= sum(pi *xi)/sum(pi)yo= sum(pi *yi)/sum(pi)

electrical loads the first step in the electrical designing for any construction is to estimate the...

Documents

electrical designing

electrical power feed

electrical load estimation

incandescent lamps

florescent lamps

small electrical devices

lighting system

living room dark colours